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Can an airplane propeller become supersonic?
Is it possible for an aircraft propeller to become supersonic?
I am not referring as for the aircraft, only the rotating blade.
I know it is extremely unlikely for a propeller plane to go supersonic, but could the propeller ?

Possible on what level? In theory you just have to spin it fast enough. Most aircraft designers in the middle 1900's knew that it was a bad idea, and tried to create designs that would avoid it.

Two examples I can think of immediately: the F-84H "Thundescreech" and the Soviet/Russian "Bear" bomber.  The former aircraft failed to enter service, in part because it was literally so loud as to make people working on the apron physically ill.  The Bear has been in service for decades, however.
Both of these aircraft had propellers that ran supersonic at the tips -- the Bear has eight of them, though they're less strident than those of the Thunderscreech (probably because slower at the tips).

I think another reason is that the Bear's props tend to do it mostly at cruise altitude where air density is less and thus sound propagation isn't as bad. I've heard reports however of pilots of intercepting NATO aircraft feeling unwell from the effect.

I seem to recall reading that the US SOSUS system (passive underwater sonar for detecting submarines) could detect Bear BOMBERS flying near them)

@DJohnM I think you're correct.

Yes the blade tips can go supersonic at redline rpm if the blades are long enough, the tip speed simply being a function of those two variables, along with the right temperature, which determines the speed of sound.
Some common airplanes have propellers that produce supersonic shock waves on takeoff, generally long two bladed ones. They produce a distinctive, very loud BLAAAAAT sort of sound as they pass by you on take off. Slowing the propeller RPM just a little makes the sound go away.  The speed of sound being temperature dependent, the effect is worse when the air is cooler than warmer.
It's common on airplanes with radials and two blade propellers like the DeHavilland Beaver. The Harvard/T6 trainer, with an R-1340 radial had a particularly intense "sonic blat" that is familiar if you've ever watched one at an airshow.
Cessna 180s/185s with two blade propellers also do it at takeoff RPM. A 185 taking off from a lake on a cool morning will have that distinctive searing sound that echoes off the local hills.
If a lake or other location is unusually noise sensitive, it's a common practice to back off on the RPM a little bit as soon as airborne, about 100 RPM or so, to make the "sonic blat" go away.
On of the benefits of switching to a 3 blade prop is lower takeoff noise, because the shorter blade tips stay subsonic.

Yes, you can figure that out yourself with simple math. The circumference of a circle is 2 Pi r, and let's say the speed of sound is 343 meters / second -- it depends on a few parameters, but 343 m / s is what the wikipedia page uses. My plane's engine redlines the rpm at 2700 rpm, that is 2700 revolutions per minute. You can now solve for r and thus determine the prop length at which the tip goes supersonic at 2700 rpm. We have:
2 Pi r 2700/60 == 343
and you get a value for r of 1.21311 meters.
You can play with the rpm, but 2700 is a typical redline for a piston plane's engine. Of course, our props aren't 1.21 meters in radius, but this equation should give you an idea on the radius / rpm relationship.

The rpm of the engine might not be the same of the propeller if there's a gearbox between them, as often is the case.

Good answer although from the question point of view probably you would start with a known tip radius and redline rpm, then calculate the speed to determine if the tip goes supersonic.  Then you can also determine if is goes transonic which I think is avoided also. Although both speeds are affected by the propeller shape, which can cause a shockwave to occur prior to the tip reaching actual supersonic speed, but that's a nuance that probably only the airplane manufacturer would get into because of the complexity.

@sophit, good minor point, I should have talked about the prop rpm, not the engine rpm.

@StevePemberton, disagree on the first part of your comment, given the question I intentionally wanted to have the radius the (only) variable to solve for. That way it's also not a question if it goes supersonic, but at which radius it goes supersonic. Your point about trans-sonic and prop shape determining a shockwave is well taken, you're right about that.

The tips of propeller blades can go supersonic, however this is very inefficient so in practice this is rarely done. One notable exception is the Russian Tu-95 turboprop bomber which cruises at 380 knots (440 mph, 710 km/h). The propeller tips of the Tu-95 move at supersonic speeds, which makes the plane extremely noisy.

Tu-95 (photo: RAF, via: Wikimedia Commons, OGL v1.0)

Yes. A former supervisor was an antiaircraft missile supervisor. He reported that TU-95 bombers were easy to spot because of the distinct signature of the supersonic speed of the propeller tips.

Welcome to [aviation.se]. You'll note that this aircraft was already mentioned in 2 other answers. Have anything else to add? Please take the [tour] to see how things work a bit differently here than at other discussion boards.

Also potentially missing here is if the signature was sound (per John K) or on some form of Radar using doppler to ID contacts which would seem more likely for an antiaircraft missile supervisor to be observing..


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Cessna 150 Battery master switches off when ignition switched on
We have a Cessna 150 with an electrical fault. When you switch the ignigtion switch to the both magneto position, electrical power is lost. Switch the ignition switch back to either the left or right magneto and the power comes back on.
If you remove the switch leads fromm the magnetos, switching the battery master on has no effect.

Hello Liam, welcome to aviation.stackexchange.com.  What is your question?

These switches typically have a fixed disc shaped circuit board with a printed circuit pattern, and a rotating element driven by the key with contacts that rub on the board and close or open circuits as the contact assembly rotates when you turn the key.
Not having the internal circuit diagram of the switch to work from, I can only make wild guesses. It almost sounds like the battery terminal on the switch is finding a ground path through the magneto P-lead or even the P-lead shielding at certain key positions and breaking contact at others. This suggests something is loose inside the switch. If you don't want to drop 200 bucks on a new switch, try to borrow one from another aircraft and see what happens.

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What is the history life of Cessna 172 airplane?
How can i find manufacture date of cessna 172P knowing serial number?

You can try the FAA's look up, the field you are looking for once you click on the N-Number is MFR Year

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Are there any sources that contain information on the airfoils of the Boeing 737-500 and 737NG Aircraft?
I have found some information about this on airfoiltools.com (http://airfoiltools.com/airfoil/details?airfoil=b737a-il) but it doesn't specify the exact version of the aircraft, as the generations have different wing sizes. Also, the site has information about a midspan airfoil, which part of the wing would it be located in? Any information about the winglets would also be appreciated.

No, this information is proprietary.
The public information about 737 foils is based on the original -100 model.
Further, it can be assumed that any modern aircraft with sufficient resources behind it utilizes custom designed airfoils that may be similar to catalog airfoils, but that are not in the catalog.


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How do you calculate the theoretical vertical speed of an aircraft?
I'm doing this for a High School research paper, so it would be preferable if little calculus is used for any calculations (up to Calculus I is probably okay). This should be based off of the flight profile of any Boeing 737 classic. The Angle of Attack is known, lift produced by the aircraft is known, as well as the airspeed (assuming that it is only in the horizontal axis). Is there any physics equation that relates these variables together?

This may help a bit -- https://aviation.stackexchange.com/a/56476/34686 -- you can see that it's not really a problem about lift. If you look at it as a vector problem, the "Thrust-Drag" vector plays a critical role. And any time you are holding angle-of-attack constant (and also not making any changes in a/c configuration like lowering/raising flaps) you can assume that the LIft and Drag coefficients are constant, which means you can pick either the Lift vector or the Drag vector and assume that airspeed must be proportional to the square root of that vector.  Well, that's just a tiny start--

(Comment above uses airspeed to mean the airspeed along the flight path, not horizontally.)

There are many answers here with the equation that you need, I like this one for example.

For a precise answer, there are secondary effects to consider.

Sorry about replying a bit late, but yes this does. Thanks for the help


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How does the ETOTS system work on the A380, and what does it limit?
To my knowledge, all A380s with Trent 900s have thrust limited to 78% N1 until 35 knots speed (not sure if ground or airspeed). Would appreciate any expert knowledge on anything else that is limited and information about the system itself.


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Why do Airbus engines seemingly spool up faster than Boeing?
When looking at the GP7000 and Trent XWB, their spool up times are incredibly fast. Compare this to the GE90 on the 777 which takes several seconds longer, why would this be the case? Though yes, the XWB has less thrust and is smaller than a GE90, they're relatively comparable. Compare this with a PW4000 which takes similar time to spool as a GE90 and they're even closer in size. Is this because of engine design, or something else?

Is that from the same idle setting? I bet the Airbus engines have higher idle RPM.

It's not as dependent on the aircraft as it is on the engine itself. For example, the Boeing 737 and Airbus A320 both use CFM56's as their power plant, and their spool-up times are relatively similar. Some variants of the 777 fly with the PW4000 equipped
as well with no significant difference in spool-up time. Airbus may simply have different requirements for their planes than Boeing in terms of engine performance and capabilities, but it is not strictly a manufacturer issue.


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Canadian PPL as an American?
I am an American hobbyist pilot who had been flying for years as a LSA pilot. Recently, I started training for my PPL and applied for medical. Sadly, it is likely my Class 3 will be rejected to past history with anxiety (long since resolved, and been off meds for years).
I live close to the Canadian border… how hard would it be to get a Canadian PPL? My cursory reading of their requirements suggests that all that is required for medical is a thorough physical (which shouldn’t be an issue for me). I have the skills and hours to checkride today if I had to.
Is this a possibility? If so, how long would it take?

Gotta love the FAA's enlightened attitude toward these issues.

Have you actually submitted the medical application to the FAA? If not, don't! If you have, there might be a way to cancel it, but I'm not sure.

@Someone As far as the FAA is concerned, you can even submit it to MedXPress and as long as you haven't started a physical with an AME you haven't started an application.

@MarcusAurelius Long-since resolved anxiety is not a deal-breaker. If you haven't seen an AME yet, do some research before doing so! Depending on the exact details of your medical records (which you certainly should not post here) it may be possible for you to be issued on the spot. But you will certainly be deferred if you go in unprepared. I highly recommend discussing your specific case with an AME before actually going into the physical. If you already have been deferred, a special issuance is a possibility. How much time and effort this would take depends again on your medical details.


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Are the details of flight control laws the same for every fly-by-wire Airbus aircraft or are they only the same within the specific family?
I was wondering if someone learns the details of every single flight control law for the A320 family, does that knowledge still apply to all the other Airbus aircrafts? If not, what are the differences?

What is FWB? Is this a technical term or from the urban dictionary? :-)

My apologies, it was meant to be FBW, which is short for fly-by-wire, but it's not clear enough, I edited it.

Even within the A320 family there are differences, for example the A318 with steep approach capabilities has modified FBW laws.

From a pilot point of view they are the same, for all families.
Technically, they are differently implemented, from a family to another one, but they are implemented the same way in the same family.
The basic principle for all Airbus FBW are as follows in normal law:

in pitch normal law the side stick send a «load factor» order that would become limited when protections are triggered such as AOA protection.
in roll the side stick gives a «roll rate» that would become limited when protections are triggered such as bank angle exceeding a threshold.
When sensors redundancy is degraded the laws become degraded and we enter alternate laws or direct law.
On ground we are in direct simulated relation ship between the side stick and the related surfaces.

The differences:
They don’t affect the flying technique from the pilot point of view, they concern the protections mostly those that concern the structure of the aircraft for exemple
The Maneuver Load Alleviation function that distributes the loads on the huge wing of the 330, this protection is specific to the 330/340

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Why were the fronts of the Messerschmitt Me-109 painted yellow?
After seeing the movie Dunkirk I learnt that the Luftwaffe didn't actually use the yellow coloring on the Me-109 until a few months after the battle of Dunkirk. Was this perhaps done to avoid friendly fire? I'm eager to hear other possible explanations.
Thanks!

I have a friend who is really interested in WWII and especially the German Luftwaffe. I asked him and he told me the yellow nose indicates the squadron leader.

You're right that the yellow tactical markings were used in order to reduce the friendly fire incidents (similar to the Allied D-Day stripes). As noted by Alan W. Hall in Messerschmidt Bf 109 warpaint:

Quick identification has always been a necessity in aerial conflict ... Channel front fighters of the Luftwaffe started carrying yellow cowl and rudder markings. Some painted the entire cowling forward of the windscreen yellow, while others just painted the undersides of the cowling.

The image below shows some of the various tactical markings used at the time.

Bf 109 tactical markings; image from asisbiz.com
While the yellow paint seems to have been used before, it saw widespread use from August 1940 during Battle of Britain, and was used in some cases in other theaters too. Also, other aircraft like FW 190 also used some variations, like the yellow painted rudder.

Thank you for your answer! Do you happen to know if yellow had any further symbolic meaning? Or was the particular choice of yellow (instead of any other color contrasting them to opposing forces) based on contrast to the blue and white sky?

JG26 pilot Steinhilper records in his diary that yellow paint arrived at aerodromes on August 24th with orders to paint all cowlings immediately.  This order was direct from Goering but unpopular with commanders. It was for identification purposes because Goering claimed that the skies would be full of Luftwaffe.  Hope that helps.

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What is the range of amplitude of vibration in a helicopter?
Vibration in helicopter can be caused by many different reasons. I am interested to know what is the range of amplitude of vibration?
I am interested to know how amplitude varies with frequency of vibration? I am looking for an answer in terms of distance from a reference or equilibrium  or stopping condition. For example does the maximum distance (amplitude) from the reference is 2 cm or less or more when helicopter moves?
I understand that it would depend on the helicopter type. So any reference to such analysis would be appreciated.
This is similar/extention to the question What is the range of frequencies of vibration in a helicopter due to the main rotor alone? where it asked about frequency of rotor.

Seems like the answers to your previous question answer this? Or at least would be a starting point?

@fooot The previous question is not the same.  I am in desperate need for some information. This site helped me (really big way) and now I need more information so i have asked. It is not that I did not search the web, but I could not find any proper answer so I am asking.

I just thought it would be good to mention your previous question here, given that it's related and also provides the frequency vs. amplitude graphs that should be relevant here.

The amplitude of the vibration depends on the frequency and the acceleration: at a given acceleration, only at low frequencies do we get any significant displacement. If displacement is taken as the intensity indicator, the frequency plot is weighted towards low frequencies - the higher frequencies have a higher energy contents, that is why often a logarithmic relative scale is used (dB).
This document gives good information and background on measuring th severity of vibration. It can be done with an accelerometer: integrating acceleration yields velocity and displacement as well. From the document:

Experience has shown that the overall RMS value of vibration velocity measured over the range 10 to 1000 Hz gives the best indication of a vibration's severity. A probable explanation is that a given velocity level corresponds to a given energy level so that vibration at low and high frequencies are equally weighted from a vibration energy point of view. In practice many machines have a reasonably flat velocity spectrum.

The vibration frequency plot in the previous answer is in dB acceleration, an earlier slide shows the reference. If it is displacement of a particular frequency in the plot you are after:

Read dB acceleration at the particular frequency
Convert into m/s$^2$ to get acceleration $a$
Amplitude = $\frac{a}{(2 \pi f)^2}$

The link to the document is not working.

Ah yeah, fixed it.


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Does the downwash velocity calculation apply to hovering flight?
After reading a very interesting interplay between two writers attempting to answer this question of helicopter downwash velocity, it occurred to me that, although this calculation:

$V_i = \sqrt{\frac{F_n}{2 \rho S}}$

may be accurate, it would not take into account the vertical velocity of the air entering the disk.
When a helicopter is in a hovering state, is it not in its own vortex?  I am not sure if the helicopter pilot can compensate by adjusting the blade AoA to the relative wind of the downflow, but in forward flight blade AoA must be changed for forward and trailing wind components to keep lift constant on either side.
Does it take a lot more power to hover?  Can we predict the velocity of the inflow into the disk to determine the outflow velocity (which should be the inflow + the calculated downwash).
We would expect lower net downwash in forward flight, but how does that compare to power required in hovering flight?

https://aviation.stackexchange.com/questions/45631/do-helicopters-use-more-fuel-when-hovering#:~:text=In%20both%20cases%2C%20the%20force,translational%20drag%20becomes%20an%20issue.

It seemed that agreeing to "split the difference" of two downwash velocity calculations required further examination.  Let's please keep working on this.

The power required for hover is plainly answered in the linked question. You may be able to rephrase your question but it needs more than just a title modification.

@Pilothead this question is about the formula applied for downwash velocity in the referenced question.  Why are you stuck on Power? (BTW, the reference provided by the moderators is excellent).  Why not take a try at an answer?

Because you are stuck on power. Your question contains "Does it take a lot more power to hover?" and "how does that compare to power required in hovering flight?". The answer to this is the question I put in the 'moderator's' link. Again, you need to do more than just modify the title.

@Pilothead is it good to be stubborn?  Please read the first paragraph and see what I am asking.  It is being suggested that increased power requirement (as evidenced by the fine reference) is required because the helicopter (in hover) is operating in its own downdraft indicating that this needs to be included in a mathematical model of predicted net downwash. (The two writers had the formula, 2x the formula, then split the difference).  While this type of. diplomatic agreement is understandable, does it have any place in science or (good) engineering?

@RobertDiGiovanni, the induced velocity is a delta, so it's w.r.t to the incoming velocity, see for example this link

@ROIMaison Yes!  Hopefully others will read the link as well.


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What are the correct Cd0 and Cdi values for a NACA 2412 airfoil?
According to an answer to this question, at best L/D the the drag coefficient is evenly split between Cd0 and Cdi.
I'm going to limit the question to the Cessna 172 and its NACA 2412 wing with an estimated Reynolds number of 2.5 million at a rotation speed of 55 knots.

Cd = Cd0 + Cdi

Cdi can be replaced with Cl$^2$/(3.14 × AR × epsilon) or Cl$^2$/K. K = around 20.
Cl values of 0.8 at 5 degrees AoA yield a Cdi of 0.032, which is indeed close to the Cd0 of 0.031.
My quandary is that if all this is true, there should be a noticeable deceleration  when the aircraft is rotated at 55 knots while taking off, resulting from a doubling of drag.  This is not observed.
Furthermore, while lift to drag ratios of the NACA 2412 airfoil are commonly well over 50 at Re > 2.5 million, the L/D (based on glide ratio) of the entire aircraft is only around 10.
Is it possible the mathematical modeling is overestimating Cdi?  How did they arrive at K = 3.14 × AR × epsilon?
Is it possible differences in Re number (when testing models in wind tunnels) has skewed the Cdi calculation too high for full scale aircraft?

Robert, at 55 knots the dynamic pressure is lower. While the induced coefficient of drag doubles when going from 65 knots to 55, the drag itself is only a bit higher and propeller thrust (being proportional to speed inverted) also is higher. The airplane will accelerate just fine when rotated at 55 knots.

@PeterKämpf yes, that is observed.  My query was why (a tricycle gear at much lower AoA) does not experience a loss of acceleration upon rotation if the calculated Cdi increases to that near Cdo (a doubling in drag).  But it may be that, as Rob McDonald and Thomas Perry pointed out, the ratio of lift generating drag compared to the form drag of the entire aircraft may not be expressed in K, which may be reflecting only the wing (airfoil)(3 dimensionally).

Answering the question:  Why are induced and parasite drag equal at best L/D?
It is easy enough to do the math...
Starting with a simple parabolic polar
$C_D=C_{D,0}+K\,{C_L}^2$
Divide both sides by $C_L$.  We will observe that extremum of D/L will occur at the same place as extremum of L/D.
$\frac{C_D}{C_L}=C_{D,0}\,{C_L}^{-1}+K\,{C_L}$
Take the derivative of both sides with respect to $C_L$.
$\frac{d}{d\,C_L}\left(\frac{C_D}{C_L}\right)=-C_{D,0}\,{C_L}^{-2}+K$
Set this equal to zero.  This condition will be true at any extremum of $D/L$.
$0=-C_{D,0}\,{C_L}^{-2}+K$
Re-arrange
$C_{D,0} = K\,{C_L}^2$
Which demonstrates that parasite drag and induced drag are equal at the extremum of $L/D$.
This is simply a property of a parabola -- that gets applied to an airplane.  It is a mathematical consequence.
We can take this some steps further...
At best $L/D$, what is the $C_L$?  Start with our last equation and solve for $C_L$
${C_L}^* = \sqrt{ \frac{C_{D,0}}{K}}$
At best $L/D$!  Note that we call this ${C_L}^*$ -- pronounced See-Ell-Star.
If we want to know the value of best $L/D$, we can divide this equation by twice the parasite drag coefficient (since we know they are equal our desired condition)
${\frac{C_L}{C_D}}_\mathrm{Best} = \left(\sqrt{ \frac{C_{D,0}}{K}}\right)/\left(2\,C_{D,0}\right)$
Square the denominator and combine with the numerator.
${\frac{L}{D}}_\mathrm{Best} = \sqrt{ \frac{C_{D,0}}{K\,4\,{C_{D,0}}^2}}$
Cancel the extra $C_{D,0}$..
${\frac{L}{D}}_\mathrm{Best} = \sqrt{ \frac{1}{K\,4\,{C_{D,0}}}}$
You can pull out the $1/4$ if you want to...
${\frac{L}{D}}_\mathrm{Best} = 0.5 \sqrt{ \frac{1}{K\,{C_{D,0}}}}$
So, for the example you cited...
$C_{D,0}=0.031$  and $K=0.054$
${\frac{L}{D}}_\mathrm{Best} = 12.22$
And, it occurs at ${C_L}^*$
${C_L}^* = 0.758$

So, it boils down to how K is derived, and what it's value will be for various aircraft types.  But thanks for this (far more mathematicly satisfying) answer.

The derivation I've given above works for any two-term parabola.  This definition of K is the conventional one.  Much of the above derivation also works for a three-term parabola $C_D=C_{D,0}+K_1,C_L+K_2,{C_L}^2$, but it gets a lot messier and some of the details don't work out the same.
As you can see, the relative values of $C_{D,0}$ and $K$ are what determine the shape of the drag polar.

An airfoil does not have $c_{d,i}$.
Induced drag is a 3D effect.  An airfoil is a 2D entity.  When you look at airfoil drag data in a book, the non-constant value (that sometimes looks like a parabola) is actually the variation of form drag with lift coefficient.  It is not induced drag.
An airfoil can readily have an L/D of 50 to 100 -- as you pointed out, the max L/D of a Cessna is about 11.
When you are rolling down the runway, you do not have zero induced drag.  The wing is at some angle of attack, it is producing some amount of lift. You then rotate to higher angle of attack (increasing the induced drag) and lift off.
The lift will increase and induced drag will increase -- but you were accelerating on the ground -- i.e. there was already an imbalance of forces, your thrust was enough greater than the sum of retarding forces that you were accelerating.
If you had not rotated (and the runway was long enough), you would have continued accelerating to higher speeds.  At some point, you might exceed the speed that the wheels and landing gear can handle.
So perhaps when you rotate, your rate of acceleration is slightly reduced.  Of course, you're also losing the rolling resistance of the tires.
The speed for rotation is determined as a speed that is fast enough to allow you to take off safely beyond the stall speed.  Immediately after takeoff, you are still in ground effect and will continue accelerating as you climb out.  When you reach pattern altitude and level off, you will be traveling much faster than your rotation speed.
In fact, although the power setting is nearly constant for this entire period, your thrust is decreasing as you go faster (prop thrust is greatest at static conditions and drops as you accelerate).  So it is clear that at rotation speed, you have thrust substantially greater than drag.

"So perhaps when you rotate, your rate of acceleration is slightly reduced".  My point exactly.  So why does it mathematicly calculate to a coefficient similar in value to Cd0?  I suspect for very slick aircraft it might, but for something draggier like a biplane, perhaps not.  Makes me think about winglets, strakes, and fairings.

No no no.  Do not use my quote against me like that.  What even is your question at this point?  Your title question makes no sense, an airfoil does not have induced drag.  Your point about sudden deceleration during takeoff was addressed by my post -- there is plenty of thrust to keep accelerating.  Do you want to know why at best L/D, an aircraft has half induced and half parasite drag?   This has nothing to do with takeoff.

Not against you at all.  The observations are in agreement.  I'm trying to figure out how the calculated Cdi is so high.

I believe I see your error.  K is usually defined as $K,{C_L}^2$.  You're defining it as $K=\pi,e,AR$.  Which means your value of about 20 should really be about .050 -- which matches.

that's 0.05 in the numerator or 20 in the denominator.  Same.  I suspect these calculations are for the wing (3D) only.

Yes, Induced drag only exists in 3D.  Airfoil drag data is one of many things that contribute to a 3D aircraft drag polar.  That said, if you were to fit a parabola to 2D airfoil data, this would still show you the best L/D for the airfoil.

Is it possible the mathematical modeling is overestimating Cdi?

Maybe they used the wrong data!
For the answer, below, a simple approach is taken that converges on the desired result. In completing some back of the envelope modeling of the 172 (in Excel), supporting background developments in my notes show the following...
We can start with the general proposition
$$T = \frac{T_{Hp}\cdot 550}{V} = \frac{V^2\rho C_DS}{2}$$
where
$$C_D = C_{D_0} + \frac{{C_L}^2}{\pi eAR} .$$
We can estimate $e$ and $C_{D_0}$ or assume a polar.
The value of $$e = \frac{1}{1+\delta} .$$
The variable coefficient $k$ in the above is $(\pi eAR)^{-1}$. Therefore,
$$C_D = C_{D_0} + k {C_L}^2 .$$
Plotting $C_D$ as y vs ${C_L}^2$ as x gives a fitted-line of slope $k$ and intercept $C_{D_0}$. Given slope $k = (\pi eAR)^{-1}$, then $e = (k\pi AR)^{-1}$.  We can do this just for the wing alone, or for the whole airplane.
But how does one do this?
Given data in the POH, the whole-airplane approach must necessarily involve something close to a PIW or VIW assessment by using power surrogates for the aircraft for which the POH data are overlaid. In the individual analysis being completed for the 172S or for the 172M, the POH evaluation for each plane was overlayed on a surrogate hodograph that could be independently adjusted for power and velocity to match the POH data.  From the surrogate, various elements were extracted as necessary. In other words, a polar-like hodograph was assumed and fitted to the corresponding measures for the aircraft. This is actually a hodographic plot of aircraft $H_p$ available (at the blade) vs $H_p$ required for flight, both as a function of velocity in knots. The analysis was completed based on fitting the derived polar to real aircraft data.  Of particular note, data were rectified to velocity in knots and elevation at sea level for standard temperature. Using this process, and an evaluation of data in the POH for each plane, some very interesting perspectives on aircraft performance can be ascertained. The results appear to be surprisingly representative or satisfactory, and are easily validated.
Here are the resulting polars -

Of particular note, data given for for the earlier version POH for the 172 M are not presented as velocities in knots.  This was noted as a possible source of error in calculations. The later 172 M POH provides velocities in knots. For the 172 M polar, above, velocities are in knots.

So at cruise of 100 knots the Cl need only be half of what it is at 70 knots and 1/4 of what it need be at 50 knots, showing Cd0 and Cdi to be dependent on airspeed.  Thanks.  Interesting the Cl$^2$ (therefor Cdi) is linear WRT Cd.

@RobertDiGiovanni Keep in mind that the induced drag is over there in that $k{C_L}^2$ thingy.  Of course $C_{D_0}$ is the drag from all that other stuff catching air... but you probably know all this already...

I suppose this is why flying wings were all the rage years ago, but modern gliders seem to be very high AR wings with smooth fairings at attachment points.

My quandary is that if all this is true, there should be a noticeable deceleration when the aircraft is rotated at 55 knots while taking off, resulting from a doubling of drag. This is not observed.

Right, this is not observed because this is not the whole story.
Just before taking off, the engine has to win the aerodynamic drag and the rolling friction of the wheels. The aerodynamic drag can be expressed via the usual $D= qS(C_{D_0}+K\,{C_L}^2)$ while the rolling friction is simply proportional to the weight of the airplane via a coefficient $\mu$ which depends on the surface of the landing field: it goes from some 0.03 for dry aspahlt to some 0.1 for unprepared wet field. So, the total force before taking off is:
$$F= qS(C_{D_0}+K\,{C_L}^2) + \mu mg$$
Just after takeoff, the rolling friction disappears and the induced drag becomes the one given by lift equals weight:
$$F= qSC_{D_0}+K\frac{(mg)²}{qS}$$
Let's do the math with the values you supplied plus the ones from Wikipedia:
$C_{D_0}=0.031$, $K=0.054$, $mg=1'111kg \times 9.81≈11'000N$, $S=16.2m²$, $C_L$ before takeoff $≈0.1$, $\mu$ of wet grass $≈0.08$ and $V=55kts≈30m/s$.
We get (just before and after takeoff respectively):
$qS≈ 8'200$
$\begin{cases}
F= 8'200(0.031+0.054 \times 0.1²) + 0.08 \times 11'000=1'138N\\
F= 8'200 \times 0.031 + 0.054\frac{11'000²}{8'200}=1'051N
\end{cases}$
As you can see there's no much of a difference between before and after rotation and actually after rotation there is a reduction of the needed thrust/power (at least in the conditions used in this example).


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Can an aircraft climb with the thrust line below the line of flight?
"Downthrust" is a common engine mount feature of propeller driven aircraft, models and full scale alike.
From model building experience and flight, downthrust, angling the (tractor) engine mount down a few degrees seems to relax static stability a bit when full power is applied by counter-acting tail down force and enabling flight at various speeds with less trimming.
The effect in a full power climb would be a higher trim speed than a power off glide at the same trim.
Yet many people on this site strongly feel that lift must be less than weight in a climb.
I think it can be (rockets do this well), but using the wing to climb seems far more efficient, as witnessed by superior climb and distance performance at Vcc (cruise climb) and Vy compared with Vx as seen in this reference
The larger slower wing produces much more force than the smaller, faster propeller.  Additionally, at Vx, increased angle of attack, with thrust used to counter additional drag may be a better strategy than reducing wing lift and trying to replace it with a vertical thrust component.  Most GA planes will lose airspeed at this point.  At cruise climb speeds, the thrust vector may well be slightly below the line of flight, yet less fuel is consumed per distance than even Vy!

any other climb path is just another point in the ratio of lift + thrust + surplus thrust for climbing

Let's apply our "surplus thrust" and see what happens:

Increasing thrust increases airspeed
Increasing airspeed causes aircraft to pitch up
Airspeed stabilizes (to trim airspeed) at higher pitch (relative to the horizon).
$cosine$ of lift vector (vertical lift) decreases

In order to maintain vertical lift we:
a) speed up
b) increase angle of attack
c) pitch up even more
Here lies the crux of the issue.  Most GA aircraft have a thrust to weight ratio of 1:4.
Attempting to replace a 300 lb weightlifter with a 75 lb one will have predictable results.

"Lift is less than weight in a climb" is based on the assumption that lift is perpendicular to the thrust line...and the (upward pointed) thrust vector is carrying part of the weight

That's not the best way to do it, and is known as "helicoptering".  the wings job is to produce enough $vertical$ $lift$ to make the aircraft "weightless", so the engine thrust only opposes drag.  It is extremely wasteful to use thrust to "lift weight", unless the aircraft is pitched at so high an angle as to make the wings drag greater than its vertical lft contribution.
a) and b) seem to be the correct answers, as seen in the "real world" as Vx, Vy, and Vcc.
Is there a way to reconcile this academic quandary?

It isn't clear to me what you are asking.  What do you mean by "line of flight"?

If you have enough thrust horsepower, you can point straight up, where the wings are making NO lift. Any other climb path is just another point in ratio of lift +thrust+surplus thrust for climb. The engine is just canted down to put the thrust line horizontal at the airplane's cruising deck angle.

I think you should add a drawing to explain what you mean.

Hard to believe people are (desperately) trying to close this.  1.  Line of flight is the direction the air craft is travelling. which, as we know from crabbing, is not necessarily where the "nose" is pointing.  2.  Ok, John, if you want to compare fuel consumption rates of rockets and aircraft, I'm game. Hint: why are helicopter rotors large and slow.  I'm hoping we can put politics aside and focus on science.  I'll be blunt, let's not cover for a design concept that perhaps needs to be scrapped.

"Yet many people on this site strongly feel that lift must be less than weight in a climb." That statement is based on the underlying assumption that the thrust line is perpendicular to the lift vector, so that the thrust carries part of the weight. Of course, if the thrust line is angled downward, many people;) will agree that this might not be the case anymore.

Thrust is best used to oppose drag. It is a misconception that a plane nose up is using thrust to "help it climb".  No, it maintains airspeed.  The lift vector must be larger as the $cosine$ of the lift vector, vertical lift, opposes weight.

<< I'll be blunt, let's not cover for a design concept that perhaps needs to be scrapped.>> well then. are you here to ask something and learn from the answers you get, or are you here to ask questions that are only a mask for lectures you want to dispense? apart from that, as Sanchises says, you better check your assumptions.

Roger, that is all I ask of others.  We can all learn something here.

Nobody is desperate to close this, and nobody has said anything political.  It just isn't a good question.

Can't resist continuing this.  Don't mind the downvotes.

Yes, it can. As you say, that lowered thrust line adds to the lift requirement, but not dramatically so.
I guess the technical term you mean is called downthrust. Together with sidethrust, this offset alignment of the propeller axis helps to reduce retrimming when thrust is changed.
Downthrust has two effects:

It increases the angle of attack on the tail slightly when thrust is increased, thus trimming the airplane for a higher speed. The angle of attack on the tail is the sum of free flow, downwash from the wing and propeller slipstream. More thrust increases the contribution of the propeller, and if that slipstream is coming at a positive angle, the resulting angle of attack on the tail is also increased, adding some lift and lowering the nose for faster flight.
It helps to shift the propeller slipstream up so the vertical tail "sees" a more symmetric slipstream. Without downthrust, the tail would fly in the upper half of the slipstream and swirl would lead to a sideslip angle there. By lifting the slipstream up, the lower vertical tail now flies in the lower part of the propeller slipstream and experiences an offsetting sideslip angle to the one hitting the upper part of the vertical.

@RobertDiGiovanni: Downthrust is meant to avoid exactly this pitch-up by adding a pitch-down moment that increases with thrust. The relaxation of static stability is a different phenomenon: The propeller acts like an additional aerodynamic surface and creates sideforces in pitch and lift in yaw. This answer covers that aspect.

Right, downthrust will create a staticly stable trim at a lower angle of attack (and higher airspeed). Why?  Naturally, the pilot (or computer) can do the same by pitching down a little.  And we do need more thrust to climb in any case.  But isn't it better to use increased thrust to increase speed slightly (Vy) or to labor away at a lower airspeed and higher AOA?

How would you design an aircraft that could go to full power at approach speed for a go around?  Like every plane ever designed, that's how.


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What should I do if I have inadvertently busted class b airspace?
I was giving an introductory flight to a friend who I am trying to get interested in obtaining his pilots license. While he had the controls, I was so focused on watching and helping him that I think I busted the bravo. It may have only been for a cpl seconds, as I realized it immediately. But, nonetheless, I still did. What steps should I take with the authorities to acknowledge this mistake?

File an ASRS report with NASA saying what you did wrong and what you should have done differently. If the FAA decides to take certificate action against you, this will be taken as evidence of a constructive mindset, and it provides immunity to certificate action, so long as this was a first offense and was inadvertent. Make sure you keep the submission confirmation number and the identity strip they send you. NASA keeps no record of who submits these reports so you will need to keep this for evidence that you submitted the report if the FAA comes knocking.
Just as importantly, learn from the experience. If you are PIC, anything your passengers do is your fault. You should be at least as vigilant when a passenger has the controls as when you do, and be certain that your proficiency is up to the task of monitoring them without losing situational awareness. It's probably prudent to only give the controls to a passenger when well-removed from busy airspace.

The recommended course of action is to submit a NASA ASRS report  (full form here) admitting your mistake and explain the situation. These are non-punitive reports and are a good sign you understand what you did, acknowledge it, accept the responsibility and are working to remediate should the FAA launch an investigation.
Depending on when this happened if you and/or the flight school has not received a phone call about it you may be in the clear but keep in mind they can be slow. Assuming your aircraft had ADSB out they would have known who you were.


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Which type of wing would be a better option for fixed-wing UAV? Mid or low?
In the context of VTOL hybrid Fixed-wing tiltrotor UAV, for the operation that would require low weight and more maneuverability for agricultural crop surveying, which option would be a good choice, mid-wing or low-wing? The high-wing can be eliminated due to its less maneuverability and doesn't offer any significant advantages(Is what I think) or enlighten me if high-wing can be a good option for the above purpose.

A tilt rotor will blow the crops down.  A lot of folks still use biplanes (or a high wing aircraft for better ground visibility).  Avoid going Belphegor as well.

I suppose that the tilt rotors will be placed on the wingtips, just like the V-22; if so, high wing is the way forward.


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What are the differences between an oblique wing and a swing wing?
How they differ from each other.
How do they work.
What is the structural differences.
How do they help in reaching mach speeds and get better fuel efficiency.
How does they both perform with max level of technology.and other such differences.

Does this answer your question?


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What kind of combustor are these engines using?
I'm familiar with the wikipedia-standard can/can-annular/annular style combustors, but I've found a design that don't seem to fit into any of these categories, used in a number of engines such as the Turbomeca Makila and Ariel, and General Electric's H-series.

It looks like an annular design but According to GE, there's no fuel nozzles... I think fuel gets sprayed in through ports on the turbine shaft? Beyond that I haven't been able to find anything, I can't even find a name to refer to it by.
I'd love to know more about this setup, like what it's called and what it's advantages and disadvantages are, i.e. why did Turbomeca/GE's engieers go with this design instead of one of the more common setups?

This is a typical engine used where space is at a premium, that explains its particular layout.
From left to right we find:

An axial compressor with one single stage; this is used as the beginning of the engine since axial compressors are more stable in respect to asymmetries in the input airflow; anyway they give a relatively small jump of pressure; that's why it is followed by
A centrifugal compressor which gives a high compression rate in a relatively small space;
They are followed by an annular combustion chamber which in this case is "curled up" giving it a "C" shape; this is again done in order to optimise the used space and reduce as much as possible its length; compressed air enters from the left through the holes, mixes up with fuel, ignites and burns towards the right; a second set of holes on the right part of the chamber let "fresh" compressed air in in order to cool down a bit the combustion products before impelling on;
The first turbine, which drives the centrifugal compressor;
A second turbine follows, driving the axial compressor;
And finally the turbine extracting the power delivered to the output shaft of the turboshaft is found, closing the engine setup.

A very concise and easy to follow description.  Thanks!

@WayneConrad: thanks

That is amazing!  What is the typical power output of this jobber?  And is there a typical aircraft application you can mention...

@ThomasPerry: the Turbomeca Arriel in the picture is used on many helicopters around the world and delivers some 700shp. According to Wikipedia 10'000 of them have been built in 50 years, i.e. 200 per year.

Once again proof that aviation. is the only decent "SE" site.  Bravo.

@Fattie: thanks  I find that many other SE are well done, especially the "scientific" ones

The combustor appears to bhe the orange part about 1/3 from the left.  It is perforated.
Flow exits the blue centrifugal compressor to the left and makes a 90 deg turn into the chamber surrounding the combustor.  Flow goes through the holes in the combustor, snakes around, and enters the first stage of the turbine.
On the top of the far side of the combustor, we see a black thing.  I think it is either the injector and/or igniter.  Likewise, there is a boss on the direct top of the case above the combustor, that looks like an access point for another 'injector'.
If this engine has four injectors, it would appear that one is out of view on the bottom of the engine and one was removed in constructing the cutaway.


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Is it possible to “buddy start” a modern jet aircraft? I was on a 737-(200?) where thrust of one jet was used to start our aircraft
A long time ago I was on a 737 ( I believe -200 but it could been -300) with an inop APU and cross bleed. We taxied out on one engine and snuggled up behind another 737 - it looked like 15 feet from our nose to the tail of the other aircraft. The lead jet spooled up on one engine and after a couple minutes and lot of shaking our no starter engined fired off.
My question is is this possible and a modern high bypass engine?

This sounds highly improbable. Where were you sitting?

I have a  very hard time imagining a scenario where APU + the ability to cross-bleed start a motor are all deferred, but the aircraft can still be dispatched for revenue service. Perhaps the OP mistook the high power setting required for an X-bleed start (much higher in the -200 than in later models) for a buddy start.

MMEL has no mention of x-bleed valve so likely it cannot be deferred.

We did buddy starts in the C-130 (practiced them more often than doing them operationally), so I'm familiar with the idea, and I've never heard any hint of the 737 having a comparable ability. For a -200 especially, with such a small fan, that sounds inconceivable. For a CFM engine, slightly less unbelievable, but to get the N2 where you need it... I strongly doubt that it could be done.
Even if it could work, sitting behind another jet blasting max takeoff power at your motor, it would be (1) outside of any published Boeing guidance, and therefore (2) reserved for a "last flight out of Saigon" scenario - not something to be done with revenue passengers on board.

I can see how a C-130 having a single spool constant speed turboshaft can be windmill started just by blowing air though the prop. A low by pass turbofan? I'd be pretty worried about FOD even if I could get the core spinning fast enough in the jet blast of the airplane ahead, and that still seems pretty implausible.

Chuck Yeager's autobiography describes buddy starting single engine jet fighters in 1950s German as a 'WWIII has started' contingency plan, but being Sabres this would have been much better geometry for this trick than the described 737, and the autobiogrpahy can be described as 'colorful' so needs to be treated with care as a source.

Yeah you could nose the airplane right into the other's tail pipe and get enough flow to windmill the back one's engine, and only a minority of the mass flow is burned so there is still plenty of oxygen.

For the doubtful, it was a 737 and the Captain even briefed us passengers on the procedure that was about to happen. i’m in the industry so I was fascinated. Again, I believe it was a late model 200 series. I too am skeptical that modern high bypass engine could be buddy started. I can’t imagine enough air could be forced down the core to bring it up to speed but I thought I would ask.

In order to start a turbofan engine, you need to spool up the core (HPC Compressor and HP Turbine) to a high enough speed to provide airflow of enough volume and pressure to sustain combustion when fuel is introduced to the combustor.  You might be able to get the low pressure rotors (fan and low pressure turbine) to rotate, but not the core to the extent to enable combustion.


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Visually see flap settings from passenger window behind wing
I've flown on a lot of Airbus and seen a lot of trip reports on YT and wonder if you can visually see the different flap settings (if you are seated at a window behind the wing), for example Up, Flap 1, Flap 1+f, Flap 2 and FULL.

I have experienced sitting at such place in different premium and low cost airlines. And I never looked something like this. So, I think this is not a very common things and could only be an airline's preference to make it easy for the aircraft mechanics and engineers.

Yes, you can definitely tell how much the flaps are deployed just by looking at the numbers on the flap itself, close to the root of the wing, which gives the angle in degrees (original picture):

I've personally seen it on many A320 but I can't tell if it's a universal thing, I'm quite sure I've never seen it on a B737 for example.

Though unless I'm terribly unobservant, I don't recall ever seeing this sort of stenciling/printing on any of the flaps I've observed on my many flights.  So I don't think this is a universal answer.

@Milwrdfan: I've personally seen it on many A320. If it's a universal thing I can't tell, I'm quite sure I've never seen it on a B737 for example. I'm going to update my answer, thanks for your comment

See also: https://www.reddit.com/r/aviation/comments/wcikbe/what_are_these_numbers_found_on_an_airbusa320/

@manassehkatz-Moving2Codidact: A couple pilots in that thread say they're for maintenance after any work on the flaps, and for a pilot to check in case of indicator failure in the cockpit.  Makes sense.

@PeterCordes: I don't think a pilot can see those markings from the cockpit. But a flight attendant can for sure.

@sophit: Correct. But on a thread of that reddit discussion, two users who I think are pilots say that cabin crew aren't trained to read those markings, so in the event of an indicator failure, one of the pilots would get up and walk back to have a look to verify their flaps state because that's safety-critical.

@PeterCordes: ah ok, that makes sense  It would be nice to know if there are some regulations about those markings.


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Multifocal IOLs under EASA
I have a question about passing EASA Class 1 medical having done an eye surgery with Multifocal IOLs installment. Was not an issue with FAA Class 1 medical as both eyes have 20/20 vision , near intermediate and distant. Have a letter from head FAA doctor that states that I am eligible for class 1 FAA medical with no limitations!
EASA says that IOLs “should” be monofocal not shall or must.
I would appreciate a good feedback or a source that would give me an answer or if someone had an experience with this issue.
Thanks in advance


text:
What is the benefit of the Digital NOTAM over the conventional ICAO NOTAM?
Us I understand it the Digital NOTAM (AIXM 5.1 Event Specification) does allow coding the events (deviation from normal operation) in a more structured form.
That allows the user to:

Link the Event implicitly to an (one or more) AIXM 5.1 Feature
Provide the information in the structured XML (AIXM 5.1) format allowing better machine processing capabilities (schema validation, filtering, sorting, ...)

Is there any other benefit?
For 1) I do see the problem, that if the Digital NOTAM is shared, the linkage is only based on UUIDs -> but if the other party is  not having the very same Static Data baseline, the UUID will not help (some kind of Natural Keys would be needed).
I miss the advantage, that I could "apply" the changes directly to the Static Data (as it is AIXM 5.1) but it seems this is not foreseen, and I would consider this as a big lacking functionality.
Further, I do ask myself how the exchange of the Digital NOTAM is envisioned to be done.
Currently the exchange of NOTAM is done as I see it in the following ways:
a) World Exchange via AFTN, where "all" NOTAM can be published (and therefore retrieved)
b) via "property" interfaces (based on open standards).
With Digital NOTAM I do not see the AFTN working any more - how should this work in the future.


text:
Incidents where airline pilots recovered effectively from errors
I am training airline instructors in competency-based assessment and grading skills, and could use examples of pilot resilience where there are both good and not-so-good competencies in play.
To this end, I am looking for examples of incidents where airline pilots made errors while managing a threat, but recovered effectively and brought the situation to a safe conclusion.
As an example, a poor decision to continue to destination despite poor weather,  effectively managed by an early go-around and diversion,
Another example might be a malfunction that was initially misunderstood and mismanaged, but later correctly recognised and dealt with.
Investigation reports, videos etc would be very helpful.

There's an entire database of crew reported incidents called the Aviation Safety Reporting System. It is a confidential, voluntary, and non-punitive system. ASRS captures confidential reports, analyzes the resulting aviation safety data, and disseminates vital information to the aviation community.
The FAA encourages crew members to report incidents so that the community can learn from them.
You can search it at ASRS.

Thanks for this.  I know about ASRs (have filed a few over the years!) but was unaware of this comprehensive database.  What a great learning resource.


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An Aircraft is a flight level 150 the DME is at sea level find maximum and minimum range?
I have taken this question from keith williams book. I know that we can find the maximum range of DME by line of sight formula which is R=1.23(Ht + Hr). But I don't know how to determine it's minimum range.

Minimum "range" meaning the lowest number that will be shown on the DME display? Note that even when the aircraft is directly over the ground station it will still take a nonzero time for a signal to get from the aircraft to the station and back, which will read as a distance.

Note also that this is a poorly-written question because flight level 150 is NOT a set distance above sea level. Fifteen thousand feet is always 15000' above sea level, but FL150 goes up and down relative to sea level depending on atmospheric conditions.

Aside from the issues others have pointed out in the comments, the method of determining the minimum DME range display is straightforward.
Since DME measures slant range, the minimum will be observed when directly above the DME antenna. So the minimum is (aircraft altitude - antenna altitude) measured in nautical miles rounded to the nearest 0.1 NM. The key point others pointed out is that both altitudes have to use the same reference.


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Why was the cone inlet design abandoned for supersonic aircraft?
The cone inlet design can be seen on the F-104 Starfighter and the SR71 Blackbird which were both Kelly Johnson projects. Also on the F1 mirage, Mig21, and so.
I understand that they were used to weaken shockwaves, but they don't seem so popular with 4th and 5th gen fighter aircraft.

Does this answer your question?

TLDR: Spike intakes were used when top speed was the most important goal in fighter design. They perform worse versus ramp intakes at subsonic speed and when maneuvering. Since the emphasis shifted to the latter, spike intakes became less attractive.
The central spike is characterized by:

undisturbed flow,
light weight and compactness, and
easy adjustability to speed and mass flow at high flight speeds by translating the spike.

Its disadvantages are:

large loss of efficiency with angle of attack and sideslip,
long internal ducting,
limited space for radar installation, and
limited mass flow at low speed.

When each new generation of fighters was designed to be faster than the last, the spike intake was the best solution. However, with the experience of the Sixties and Seventies it became clear that a conflict when much time is spent at Mach 2 will not happen. Instead, maneuverability at high subsonic speed became the design driver, initially without giving up the Mach 2+ capability (late Sixties and early Seventies - F14, F15, Su-27, MiG-29), so rectangular intakes with moveable ramps replaced spike intakes. Later, even the Mach 2 capability was dropped (since the late Seventies - F5, F16, F18, Rafale) and simple pitot intakes took over. Better simulation also helped to understand intake flows better and optimize performance: Both the F16 and Rafale intakes are placed such that at high angle of attack the airplane structure helps to get more and better pre-compressed air to the intake, a clear advantage over a central spike intake at speeds up to Mach 1.6.

Thank you so much. This clears that up. I was standing next to a Mirage F1 yesterday and wondered why such an inlet design was no longer popular. It isn't too long to read, I read everything. Thanks again for taking the time to type this up.

All high speed supersonic inlets are designed to control and shape the shock waves.
Slower supersonic aircraft -- like the F-16, have a pitot inlet.  This forms a single normal shock in front of the inlet.
As you go faster, you use multiple shocks (the final shock is always a normal shock).  First, one external oblique shock followed by the normal.  This is an external compression inlet.
Then, you go to multiple external shocks before the normal.
Then, you use both external and internal shocks.  This is a mixed compression inlet.
Each of these stages has an increase in complexity.  In order to get these inlets to work across a range of Mach numbers, they need a series of movable ramps that must be controlled.  They often involve boundary layer suction to prevent massive separation in the inlet.  Consequently, increasing the top speed required of the aircraft drives complexity and cost of the inlet system.
When designing the shock system, you can have a 2D system or an axisymmetric system.  The 2D system uses 2D oblique shocks (ramps) to form the inlet -- this is what you see in a F-15, F-14, and most of the aircraft you mention.  The axisymmetric system uses conical shocks to form the inlet.
For the most part, you can achieve similar performance with a 2D or axisymmetric system.  The choice is more about how it integrates with the rest of the aircraft.
The axisymmetric system has less design freedom -- for example, you can't tuck an axisymmetric inlet under a wing.
The design of the latest generation of fighters is dominated by stealth considerations -- so if you're asking why they look different from previous generations, the answer is almost always stealth.
It would seem that integrating a 2D inlet into a stealthy configuration is easier than doing so with an axisymmetric inlet.


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Tecnam twin p2006t electrical master switch when towing
Does anyone know why the electrical master switch must be on before towing the p2006t? Seems odd to me; there is a reason, I just can't find it. Who knows this answer? Peter.

Presumably, you want to have the rotating beacon (or equivalent red top/bottom strobe) on while the aircraft is moving, whether while being towed or moving under its own power. There may be other reasons specific to this particular aircraft, but as a first thought, if not quite a definitive answer, I'll suggest that. Does the checklist that has you turn on the master switch specify other steps that would need electric power?


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How is true altitude derived?
I read everywhere that the true altitude is computed as:
True Altitude = Indicated Altitude + (ISA Deviation × 4/1000 × Indicated Altitude)
But why is that the case? How is that derived? Where does that 4/1000 come from?

The exact formula to compute true altitude is:
$$h_{true} = h + \frac{h}{T_0}\cdot\left(T_\mathrm{OAT}-T_\mathrm{ISA}\right)\tag{1}$$
where you can see that the value usually represented by $\frac{4}{1000}$ is in fact $\frac{1}{T_0}=\frac{1}{288.15}\approx0.00347$. You can check how it's derived here

It is not an exact formula. Taking 288.15 K is a cruder approximation, but taking one value is still an approximation.

For more see, e.g., The Hydrostatic Equation in the Evaluation Algorithm for Radiosonde Data Your link also correctly notes that it is not truly a "true" altitude.

I say it's exact because in order to derive it, you don't need to take any shortcut (approximations, Taylor expansions or similar). It can be derived directly from the barometric equation and the definition of the ISA atmosphere. But the ISA itself is an approximation so in that sense, you're right, it's not exact: despite the name, it doesn't really represent the true altitude.

It isn't. Indicated altitude is a function of just the outside pressure and the altimeter setting. The pressure at a given absolute altitude is given by:
$$ P=P_0\exp\left(\int^{\rm altitude}_0-\frac{g}{R(h)T(h)}{\rm d}h\right) $$
where $P_0$ is the ground-level pressure, $R$ is the real gas constant for air (which depends on humidity) and $T$ is the at a given height AGL.
You don't have to evaluate this integral: the point here is that it depends on the temperature and humidity at every altitude between you and the ground. In other words, the relationship between pressure and absolute/true altitude depends on not only the local outside air temperature, but also the temperature lapse rate and the humidity at every level of the atmosphere below you.
The formula you've quoted is an approximation that works when the atmosphere is "kinda standard," i.e. when the temperature is non-standard but the temperature lapse rate and humidity are standard. It is done by evaluating an integral like this, given specific assumptions about the temperature profile, and then approximating the resultant equation as a straight line.
In short, it is impossible to compute your true or absolute altitude using only barometric information, outside air temperature, and altimeter setting. Having an altimeter setting and temperature as measured at a nearby airport helps a lot, but it isn't sufficient to determine the true altitude unless your plane is sitting on the ground at said airport.

Yes this is the "truly true altitude", but it's impossible to compute using a barometer.

"it is impossible to compute your true or absolute altitude using only information available in the plane.". Well, that's not entirely true. The GPS can give you that information.

@fab Yes, that's my point. I've added a paragraph to make that more clear. Any formula picked using only local information can be dramatically wrong under unusual meteological conditions.

@fab Fair point. (Though there are a lot of weird subtleties to GPS-derived altitude). Is this wording better?

Yes, that's 100% correct now.

authors: Vladimir F Героям слава


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Identify the Albatros in this photo
I posted this question in History SE: https://history.stackexchange.com/q/71705/60022.  And it was suggested that I post a related question here.
I'm a member of Austin Sängerrunde which is a German singing club founded in 1879.  This photo hangs in our club room.   It's supposedly the Red Baron and the inscription is a greeting from him to our club.  I was told 20 years ago that the photo was a hoax.  The History SE answers make it almost certain that the handwriting is not the Red Baron's.
But perhaps it really is a photo of the Red Baron which our prankster wrote on.
We'd like to run down as much information as we can, so my question here is, "Can anyone identify the particular biplane in this photo?"  There are letters and numbers on this plane.  No picture in Google Images shows a Red Baron plane with numbers on the side.  I'm hoping someone here knows what the numbers mean.  No doubt the "Alb" is for "Albatros".

That's an Albatros CIII (right there on the side), the Red Baron flew lots of types but none of them a CIII. I cant find any info on serial number 2143 though :(

Is it beyond the realm of possibility that the Red Baron simply had his picture taken in front of someone else's plane?  Who are those other 2 guys?  Maybe it belongs to one of them...

Yeah I don't think Richtofen ever flew observer aircraft.

The Albatros C.III model was produced by half a dozen different companies. From what I could find on the internet, "Alb" was the official abbreviation used to indicate a plane built by Albatros Flugzeugwerke GmbH. The serial numbers appear to have comprised two parts, with the "suffix" usually separated by a slash, although a dot is used in this case: ".16". Perusing photos and drawings online, I could find planes using /15, /16, and /17 for the suffix, whose meaning I have not been able to establish.

@njuffa  Might 16 be the year 1916?

@B. Goddard The same thought popped into my mind. If this plane was built in 1916, it is less likely to be von Richthofen's, because according to the list of his aerial victories, he was flying other types of planes by fall of 1916.

Alb is indeed for Albatros. The German serial number system was formalised in 1913, and gives some information on identifying this plane:
Alb - Albatros Werke AG, the manufacturer, founded in 1909. Although you'll sometimes see a plane referred to as "an Albatros(s)" that's rather like saying "a Messerschmitt" or "a Boeing", and doesn't indicate the type (e.g. Richthofen himself speaks of flying "my good Albatros"). During WW1 Albatros manufactured over 10,000 aircraft of various types.
C - Two-seat armed biplane scout. There were many other C planes from various manufacturers, and the pilots themselves often referred to their planes using this letter designation. Richthofen himself talks of flying a "C-type" plane in Russia in 1916, possibly a C.III.
III - the manufacturer type in Roman numerals.
Thus to identify a German WW1 plane, you need all three parts. There were, for instance, other Albatros C planes (I - XV); there were other manufacturers' C.III planes (e.g AEG, Halberstadt), and there are other Albatros III planes (e.g. the famous Albatros D.III, where D indicates single-seat fighter). This plane is, therefore, an Albatros C.III .
2143 - the manufacturer serial number
16 - the year the aircraft was ordered (1916)
The Albatros C.III was one of the most important aircraft of the mid-war, one of the first two-seaters with weapons for both crew members. So popular was it that production was spread between Albatros and seven other manufacturers. The aircraft in your picture was built under license by Bayerische Flugzeugwerke, from a batch of 50 ordered in July 1916 (serials 2124 to 2173).
I would speculate that, because this aircraft has no squadron markings it is either about to leave its factory, or just arrived with its squadron. It is also lacking the observer's machine gun. Certainly it's unusual to see a plane in this state, with such a prominent manufacturer's mark and no other markings.  So that would make the most likely date for the photo as mid 1916.
I can't say who is in your picture, but here is the Red Baron in front of the same type of aircraft in 1917 (it's definitely not the aircraft in your picture, however, and nor was he flying it: he was just taking a ferry-flight.)

This wikipedia page:
https://en.wikipedia.org/wiki/Manfred_von_Richthofen
says this regarding his early flying career:
"The next month, Manfred joined Kampfgeschwader 2 ("No. 2 Bomber Squadron") flying a two-seater Albatros C.III."

Google finds multiple books predating Wikipedia and the Word Wide Web that mention von Richthofen flying this type of plane. Example: Edward Jablonski, The Knighted Skies: A Pictorial History of World War I in the Air, Putnam 1964, p. 78: "The Albatros C-III, a two seater reconnaissance aircraft, was widely used from its introduction in 1916 to the end of the war. Several famed German pilots began their careers flying this plane, among them Manfred von Richthofen, Ernst Udet, and Hermann Goering."


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soot on a plane
This is the nacelle (?? I guess, I'm just someone who takes plane trips ??) of a DeHavilland Dash 8-400 that I flew on (as a customer) from Toronto to Boston last week. I noticed the soot around the bottom of the exhaust, which seemed unusual to me; I'm guessing that this is the result of the smoke in the air from Quebec wildfires (this plane is owned by Porter Airlines and had presumably been operating in the northeastern US and southeastern Canada for the previous few days).
A few closely related questions:

Is this soot build-up unusual, or is it typically there and I just haven't noticed?
If it's unusual, is it likely a result of operating in a smoky environment?
Does operating in wildfire smoke impose an extra maintenance/cleanup requirements for commercial aircraft?

This is pretty standard build up and has nothing to do with operating in and/or around wildfires. You can see identical build up here, here, and here. What you are looking at is the engine exhaust and is a prime location for exhaust/soot/oil/grime buildup.
Smaller piston planes are particularly notorious for this kind of build up.


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How do I estimating night time and cross country time for thousands of flight hours?
After flying many years for a major airline I am faced with the task of estimating my night time and cross country time. I have over 16,000 hours. I need this information for my resume as I am planning to fly past 65 years of age after flying with a part 121 carrier. My company online flight log does no break out night time and cross country time.

Any Part 135 company should realize that you have quite a bit of night and cross country time based on the number of hours flown at a 121 airline.  I wouldn't worry about tying to estimate anything.

I would just make an educated guess, with a conservative number that makes sense, and put that in my CV, possibly with "estimated".  If an employer asks, explain the truth. I would expect that unless you are a freight dog it's probably 8:1 day/night. Or poll your mates who do record that those numbers, average them, and use that. Nobody's going to audit your log book.


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Two Engines Powering One Prop via a Shared Pulley
This question is specifically in reference to Paramotoring. I've seen several videos about paramotors powered by the RC Engine DLE 170, and in all of them it struggled to spin the prop at it's peak 7500 rpm, instead being limited by power to about 5-6k rpm. So my idea was putting two DLE 170s in sequence and combining the power via  shared pulley, at a rate of 1:1.3 or 1:1.25 to put both motors in their powerband and get the full potential out of them. Would this even work? I haven't seen evidence of anyone else running this setup, which usually means there's something wrong with the idea. In theory it would weigh similar to a Moster 185 and have a slightly higher output, assuming there isn't an excessive efficiency loss in the design.

"assuming there isn't an excessive efficiency loss in the design." Your assumption will rapidly proven wrong ...


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Lost Pilot logbooks/flight time records
I hold an ATP certificate Multi-engine land, commercial privileges single engine land. It was issued in 1973 just prior to leaving the Army.  As I understand it, the facility that held the Army flight records experienced a fire some years ago and the records were lost. Now, I've also lost my log books and the only thing I have to show I can fly is my ATP.
I have approximately 3,100 hours total time. What do I do?

You can use the times from the last 8710 as proof of those hours.  The examiner is required to validate those hours before signing the application.


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Why phase lag on helicopter blade is sometimes 90°, and sometimes less?
On semi rigid rotor head, phase lag is 90°, and also the advance angle between pitch link and coresponding blade. But of fully articulated rotor the phase lag is less then 90°, and also the advance angle.

Wikipedia proposes an answer.

How much are you actually comfortable with the concepts of mechanical oscillation, natural frequency and resonance? Just to tailor the answer.

I enclose some text from 'Principles of Helicopter Aerodynamics' by Gordon Leishman, on the subject:

It would be better to copy the text here, in case the links go stale.

Right... Done...

I'm not sure there's a clear, intuitive answer to this, you'll have to consider the math to some extent.  I've attempted to explain that below.  In short, a fully articulated rotor has a flap hinge offset from the center of the hub which effectively "stiffens" the blade in flap dynamics and increases flap frequency.  This causes peak flapping to occur more quickly after an applied force/moment.
A semirigid rotor flaps "freely" about a hinge at the center of the hub.  The force limiting flapping is the component of the centrifugal force perpendicular to the blade.  The centrifugal force CF pushes away from the axis of rotation/shaft.
When a blade is at 0 flapping angle the CF just stretches the blade outward, it does not push it up/down in flapping.  However, when a blade flaps up to an angle $\beta$ a component of the CF pushes it blade back down: $CF*\sin\beta$.  Vice versa when it flaps down.
I'll omit details of the math because you can find that here:
Basic Flap Dynamics.  In short, with rotor speed $\Omega$, this results in a differential equation $\ddot{\beta}=-\Omega^2 \beta$ whose solution is a 1/rev flapping frequency (it's just like a simple spring/mass system).  Furthermore, adding a flap moment to this equation at a particular azimuth results in a peak flap angle 90 degrees later.
The math is different for a fully articulated rotor, because the flap hinge is offset from the center of the hub.  Moving the flap hinge outward decreases both the flap inertia of a blade and the moment due to CF.  However, it decreases the inertia more than the CF and therefore increases flap acceleration and "stiffens" the blade in flap motion.  The differential equation therefore results in a higher frequency solution.
Again you can see the details here.  Since there's more acceleration toward 0 flapping, the flap response to an external force/moment will generally be smaller and occur sooner, within 90 degrees.  The precise number depends on how far the hinge is from the center of the hub.


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H145 helicopter type rating trainings
There are different variants of H145 helicopter (BK117), there are BK117, BK117 C2, BK117 D2 and BK117 D3 helicopter. So if you finish type rating training for example for BK117 D2, and you want to fly D3 variant, do you need any kind of flight training (differences training), or only theoretical (familiarization training) is enough, and where can I find more informations about this.

Between D2 and D3 you don't need any differences training, they are very similar.


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I need help with this dent classification rule
I'm an Aerospace Engineer from Chile, and I have the following reference issue:
Supposedly there's a damage rule related to dents on aircraft skin that says

If the distance between two dents is lower than twice of the biggest
diameter of the greater dent, they could be considered as one dent

, but I can't find the norm that references it.
Could anyone help me to figure it out?

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer.

I can't find anything in FAA reference documents that talks about dent spacing...


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Separation between aircraft changing flight level and aircraft flying on the same track at that flight level
I've read in "Optimization of aircraft trajectories over the North Atlantic Airspace" and saw that in the separation standards, the longitudinal separation when changing the flight level is 2.2 minutes. I have not seen this type of separtion before. Can someone tell me where did they get this separation from? And how did they calculate the value "2.2"?

There is no surveillance radar in oceanic airspace to track aircraft for ATC. This means that we need a way to separate aircraft without knowing exactly where they are at all times. One way to do this is with time-based separation. This means that the lead aircraft will provide periodic position reports and estimates of when they will reach their next waypoint. The aircraft behind them will be instructed to not cross the same points until a certain amount of time after the lead aircraft passed those points. This delay ensures separation between the aircraft. The aircraft are also instructed to fly a specific Mach number as part of their oceanic clearance, so if both aircraft are at the same Mach number the separation will remain fairly constant.
The separation standards you're referencing here are specific to the thesis. The beginning of the paper discusses how lateral separation (between separate tracks) was reduced from 60 NM to 25 NM based on the adoption of new surveillance and communication technologies. They then apply this to aircraft on the same track:

...reducing 25 NM to 20 NM is conceivable since we are considering
consecutive flights in the same track.

And then based on an aircraft speed of 600 kts, it would take 2 minutes to fly 20 NM, therefore:

...we assume that the longitudinal separation of consecutive aircraft
following the same track can be reduced from 10 minutes to 2
minutes...

And then when discussing flight level changes in 3.2.5, they reference that previous logic:

The separation standards, in this case, become 2.2 minutes (refer to
Section 1.2.3).

The justification seems to be that the 10% increase is less than the 50% increase in separation when changing tracks because flight levels are much closer together than tracks. But it's not clear to me where exactly they got the 2.2 minutes from.
Since this paper was written, the North Atlantic separation standards were reduced to 14 NM longitudinally and 19 NM laterally.


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"KC" designation for tanker aircraft
The names of modern tankers in use by the USAF (and some other air forces) generally seem to have a "KC" prefix, e.g. KC-135 Stratotanker, KC-46 Pegasus, KC-767, and so forth.

Does the "K" in "KC" stand for anything specific? According to this Air Force publication regarding designating and naming defense military aerospace vehicles (Attachment 3, page 24), "K" is used for tankers, but why? Is it just an arbitrary non-reserved letter, considering alternatives like "T" and "R" are already taken?
According to the same publication, "C" stands for cargo, so "KC" would be for tanker/cargo aircraft. Are there any tankers that are not also cargo?

Did you try googling this before asking?
First hit on google 'KC designation aircraft'

@ROIMaison I did but couldn't find any useful results with the terms I used.

Interesting - According to @ROIMaison's link, "K" for Kerosene and "C" for Cargo. i.e. it's a cargo transport dedicated to one very specific cargo.

Please post that "K = 'kerosene'" as an answer. Wasn't aware of that before.

With regard to the #2 question, yes: when the U.S. Navy used the A-6 in a tanker role for other carrier-based aircraft, it was designated the KA-6D.
Other carrier-based attack aircraft like the A-3 seem to have had similar designations.

The “KC” means Kerosene Cargo. (See comments section for source)
I did some research and I don’t think there’s any U.S. tanker name that doesn’t have the K on it.
Edit : I’ll just add the link here for convenience.

The DOD specification DoD 4120.15-L does not support your claim. It simply says K = Tanker and C = Transport.

@OrganicMarble, Hmm interesting. See the third comment  in the original question. Is that explaining something different?


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Were there height requirements for SR-71 pilots?
I recently sat in the SR-71 cockpit at the Museum of Flight in Seattle. I am 6'5" tall and it seemed I was too tall (with the existing seat configuration). Another post on this forum noted that the seat height could be adjusted. So, were there minimum and maximum height requirements for the pilots?

Yes, there were height requirements for pilots in the USAF. I can say that back in the 70's when I went in (and I don't think they had changed much in the previous 20 years or so) there was a minimum height of 64" and a maximum sitting height of 39". Sitting height was a bigger factor than overall height. I knew a few 6'4" and 6'5" pilots that made it due to long legs and a shorter sitting height. It was possible to get a waiver if you didn't meet these requirements.
In 2020, the AF removed initial height requirement for officer aviators.
They have instead shifted to "an anthropometric screening process to individual applicants for placement in an aircraft they can safely fly as they pursue a rated track."
Eliminating arbitrary standards will help with recruiting, especially when the number of otherwise qualified recruits is dropping. My gripe at the time was the 20/20 vision standard. Once you were in the pilot pipeline, if you needed glasses it was no big deal. But you had to have 20/20 to get in.
From the linked press release:

While still preserving safety of flight, the policy adjustment
prevents initial applicants who are below 64 inches or above 77 inches
in height from requiring an accessions waiver.
“We’re really focused on identifying and eliminating barriers to serve
in the Air Force,” said Gwendolyn DeFilippi, assistant deputy chief of
staff for manpower, personnel and services.  DeFilippi, who chairs the
Department of the Air Force Barrier Analysis Working Group, explained,
“This is a huge win, especially for women and minorities of smaller
stature who previously may have assumed they weren’t qualified to join
our team.”
With the removal of the blanket height standard, the medical and
operations communities will apply an anthropometric screening process
to individual applicants for placement in an aircraft they can safely
fly as they pursue a rated track.
“Studies have shown that women’s perceptions about being fully
qualified for a job makes them less likely to apply, even though there
is a waiver option.” said Lt. Col. Jessica Ruttenber, Air Force
mobility planner and programmer and team leader on the Women’s
Initiative Team, who led the height standards adjustment effort.
“Modifying the height standard allows the Air Force to accommodate a
larger and more diverse rated applicant pool within existing aircraft
constraints.”
Under the previous Medical Standards Directory requirement, the height
requirement to become an Air Force pilot was a standing height of 64
inches to 77 inches (5’4” to 6’5”) and a sitting height of 34 - 40
inches. Although most height waivers were approved, the previous
restriction eliminated approximately 44 percent of the U.S. female
population between the ages 20 to 29.

Would those requirements apply to, for example, KC-46 pilots?  I guess they have to fit in a trainer even if the combat job has a spacious cockpit.

The air force doesn't recruit people for a particular plane. It wants people who can, at least physically, fit in any plane. (When I was trying to join I was told "we don't want people who want to fly cargo planes. We want people who want to fly fighters, because that's the hardest people to find. There are plenty of people who want to fly fighters but can't, and we put them in the cargo planes". Might have been just the one guy I was talking to though.)

@AnonymousPhysicist  It would have in my day, but not under the current guidelines. See edited response.


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Can a B-707 recover from a spin?
There was an incident in 1959 involving a Boeing 707 on a training flight near Paris, France where a badly executed MCA demo reportedly resulted in the plane entering a spin, had an engine torn off the aircraft, and then recovered to level flight.  This is described in https://aviation-safety.net/asndb/334056, which in turn cites a UPI report reprinted in the New York Times.  The article is https://nyti.ms/3V26l0I which is sadly behind the NYT paywall, but the relevant quote is:

A Pan American World Airways spokesman said the right outboard engine
had torn loose when the craft went into a sudden spin during a
deliberate demonstration of minimum control speed.  He said the pilot
had righted the plane immediately at 8,000 feet and had flown to
Lon[d]on on three engines with no difficulty.

I'm having trouble believing a 707 can recover from a spin.  A wing drop, sure, but recovering from a fully developed spin seems implausible.  Unfortunately, I've been unable to find anything that doesn't just refer back to this UPI article.  Anybody know anything more about this incident?
I'm also astounded that the crew would fly 230 miles back to London instead of landing at the nearest runway after having an engine torn off the plane, but maybe that's just how pilots did things in 1959.

egad! That aircraft came to a grizzly end too https://en.wikipedia.org/wiki/Pan_Am_Flight_214

A spokesman for an airline might say the recognizable "spin" instead of the precise but unrecognizable jargon "spiral dive" which is easier to recover from.  Answers to https://aviation.stackexchange.com/q/753/31425 confirm your hunch that the 707 couldn't recover from a true spin.  Welcome to Aviation SE!


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Why B737 CFM56 engine sounds like heavy wind sound at Throttling up?

14:14 ~ 14:21 : Throttle Up (with heavy wind sound
14:22 ~ 14:23 : Engine reached certain %, and no heavy wind sound
14:23 ~ : Throttle Up (with heavy wind sound)
I always wonder why only B737NG with CFM56 Engine sounds with heavy wind sound when they throttling up. Once the engine reaches a certain percentage, the wind sound disappears, but is this simply because of the sudden suction of air due to throttle up? or is there another reason?

This is a great question! I hadn't noticed that before. I've checked a few other 737NG takeoff videos, and it seems like if you are sitting near the exhaust (not intake) you can definitely hear this. I wonder if it may have something to do with how the FADEC spools the engine when the TO/GA buttons are pressed on the throttle. If you watch this https://www.youtube.com/watch?v=2Ve89DuLFwY at 5:37 you can see the 1st officer click those buttons, and the lever advance then pauses for a second at 5:41...? The timings seem close to the ones in your video

@AlastairWyllie "lever advance then pauses for a second at 5:41" That is the SOP, also when not using the AT. You always let N1 stabilize at around 80% to make sure thrust is symmetric and let both engines spool at the same time.

@DeepSpace - From what I understand often one engine will inherently spool faster than the other, and the symmetric thrust can be problematic if going straight from idle to 100%. By holding at 80% you let the slower engine "catch up", and then while there will be some asymmetry going from 80 to100% it's not enough to create a problem.

Not 80%, but 40% N1. Get both N1s stabilized at 40%, then click TOGA and let the AT advance power to the computed takeoff thrust. The TOGA function itself has no such pause built in.


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EASA Regulations regarding Private Aircraft for PPL training
Can somebody advise what are the EASA specific regulations that apply regarding the use of a privately registered aircraft for PPL training? Links in particular to the EASA website would be greatly helpful.
Many thanks!

mind that ALL aircraft not registered to government agencies are privately registered. So if privately registered aircraft weren't allowed to be used for flight training, only the government would be able to provide flight training!

Each country participating to EASA has its own air authorities and local rules. Some rules are already common at EASA level, found in the Standardized European Rules of the Air, SERA.
As an example of national rules: For France, rules are edicted by DGAC, and published by SIA, the ICAO AIP manager. The link between EASA and French rules is described in this SIA document (in French). A consolidated document, SERA Complet includes both SERA and local rules. This structure may be different in other EASA State members. EASA is not the same geographic scope as EU.
SERA and related documents all related documents are named "Easy-something", but their structure with endless links is confusing and they are difficult to use due to their bureaucratic jargon, nothing in common with US CFR. Still, my gest guest is the aircraft requirements for PPL training are found in:

Easy Access Rules for Organisation Requirements for Aircrew (Part-ORA), in particular AMC1 ORA.ATO.135 Training aircraft and FSTDs.

Easy Access Rules for Declared Training Organisations (Part-DTO), in particular DTO.GEN.240 Training aircraft and FSTDs and following paragraphs.

Acceptable Means of Compliance and Guidance Material to Part-FCL.

EASA has a homepage for GA, I suggest to join their community and ask there for details.


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what is best approach to make high thrust from nozzle side? higher velocity or higher pressure?
Producing high thrust form jet engine is not one step. Actually, there are many factors that can significantly play rule here.
Anyway, am focusing only on the nozzle side and its rule only in my question and excluding all other parameters.
So, after the hot gases leave the high pressure turbine, in order to make highest thrust from the nozzle, what should we do? maintain higher pressure or higher velocity?

The answer to this question depends on whether you are considering an engine for a supersonic high performance aircraft -- or a subsonic aircraft that requires good fuel efficiency.
A subsonic aircraft will basically always have an exhaust that is subsonic at the nozzle exit.  In this situation, the nozzle exit pressure will match the ambient pressure.  This is called a fully expanded condition.  Matching this exit condition will actually determine the mass flow through the engine.
A supersonic aircraft will likely have an exhaust that is supersonic at the nozzle exit.  In this situation, the nozzle exit pressure will be either over-expanded (Pe<P0), under expanded (Pe>P0), or fully expanded (Pe=P0).  Under and over expanded nozzles will set up a pattern of shock diamonds, starting with either shocks or an expansion fan.  The fully expanded condition should not have diamonds in the exhaust.  In this case, flow will be choked at the nozzle throat -- which sets the mass flow through the engine.
The thrust equation for a jet is something like:
$Fn=\dot{m}\,(V_j-V_i)+(P_e-P_0)A_e$
You will get the best thrust when you are fully expanded ($P_e=P_0$) and therefore focus on $\Delta V$ -- i.e. maximize your $V_j$.

“A supersonic aircraft will likely have an exhaust that is supersonic at the nozzle exit” - Do you mean with an afterburner? I read somewhere that the speed of a jet exhaust is most of the time subsonic, unless there is an afterburner.

In order to have positive thrust, Vj > Vi.  So if the aircraft is supersonic, the exhaust must be too.  So we're left with the case of a supersonic capable aircraft that is flying subsonically at the time...  So, depending on the flight condition and most importantly the throttle level, it could be either way.  With AB engaged, you will have supersonic exhaust (even when flying subsonically).  But at part-power, you could have subsonic exhaust in a supersonic capable engine.

I see. So aircraft that are able to supercruise have supersonic exhaust speeds without the AB engaged? Interesting, thanks.


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How could jet engine spools get connected?
I understand that the design jet engines have two or three spools in order to make every component running within its optimal design. But that's still not clear to me.
So, a bit of confusion still exist here. If you could pls explain how they are connected together?
Are they using coupling?
OR
they extend spools inside each other?
Thank you

In short
A spool holds the spinning compressor (or fan) rotors and the turbine rotors. Each spool is independent of the other(s). Still all spools must have their shafts centered within the engine, i.e. within the stationary vanes and the outer duct, and they must be at a fixed location on the longitudinal axis.
Centering and longitudinal lock are done using two stationary frames fixed to the wing, and bearings. One of the shaft, the N1/low pressure spool is solid, N2 and N3 shafts are hollow, and coaxial to the main shaft, at a distance from N1 determined by bearings. The frames keep all bearings (and the rest of the engine!) at a fixed position relatively to the wing.
The frames are often confused with stator vanes. One of the frames, the fan frame on the left, along with the cage and inner ring of a bearing (frame and bearing are not a the same scale):

Fan frame and shaft bearing. Sources: left, right.
Details
The slower spool shaft (N1) is usually directly supported by the frames through bearings. The other spool shafts are supported by the N1 shaft and other bearings. There are variants, e.g. for a CFM56-7B with two spools:

CFM56-7B architecture, from CFM56-7B familiarization manual.
Bearings are magenta disks (ball bearings) and squares (cylindrical bearings). The green N1 shaft is supported by the two stationary red frames. You can see the blue hollow N2 shaft is coaxial to N1, centered by two more bearings, one with the fan frame, one with N1. This results in this structure:

The pink parts are the two conical ends of the an incompletely shown N2 shaft (blue on the first diagram), the golden shaft is N1 (green on the first diagram).
Cylindrical bearings are used to center the shaft within the frames while ball bearings are used both for centering and to prevent shaft longitudinal motion.
Both spool are spun by their respective turbine on the right side. Each spool spins a compressor (or fan) on the left side. It may be easier to see this in motion on this video:

The specific N1 and N2 rpm are determined by the balance between the different sections of a spool and the different spools. CMFI uses 2 spools, Rolls-Royce uses 3 spools. Each design has pros and cons. RR is able to get the different flows and rpm they want more easily than CFMI, at the price of a more complex shaft design. On the other hand CFMI has to adjust the less than optimal resulting flows and rpm using vanes with a variable angle of attack (VSV), bleed valves, etc, which also add complexity.
More:

How are engines mounted onto wings?, explains how the two frames are fixed to the wing.

In a turbofan what holds the spinning axis?, gives more details about the mechanical design of the spools and the frames.


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Why does Air Force Two lack a tail number?
Some people have been claiming photo manipulation on shots of a recent political event in Detroit. I don't believe the claims, but one thing was brought up that was confusing: The C-32A (I assume) that's serving as Air Force Two at this event doesn't appear to have a tail number.
This is a screenshot from a video posted by an attendee at the event.

I would have expected to see the tail number there, as seen here:

I know they built 4 C-32As, and I assume they're all still in service as Air Force Two, but I would have expected all of them to have the tail number on both sides of the tail. I thought that was a legal requirement. Anyone know what's going on?

 US Air Mobility Command Has Ordered The Removal Of Numerous Tail Numbers In Recent Years
Air Force One and Air Force Two are both included in the order which otherwise involves cargo and refueling aircraft.
The reasoning given by USAMC is operational security.

Amazing. Bad guys telling apart the two VC-25 Air Force One planes was literally the plot of the pilot episode of NCIS.  20 years later…

How do these show on flight traffic displays as a type class or flight number for the specific duration of an operation, a random number?